Why are the derivatives of trigonometric functions again trigonometric? The derivative of a periodic function would of course be periodic, but I am just curious to know why the derivative of a trigonometric function does not go out of the trigonometric world?
 A: So to understand that, one has to understand how the trigonometric functions are defined. Perhaps the most important property is
$$
\sin^2 + \cos^2 = 1
$$
which tells that we are looking at the coordinates of points on the circle. However, there are a lot of ways to go over the points of the circle. For example taking $\sin(x^2)$ and $\cos(x^2)$ would work as well. Or we could also count in an other unit than radian, which would also change the derivative. This tells us that the other condition to define $\sin$ is the condition to go over the points of the circle at speed $1$ (The value of the angle corresponds to the arclength of the portion of circle). So, the derivative of the vector $u_x = (\sin x,\cos x)$ has to be again an element of the unit circle, and so is expressed with the same trigonometric functions (it is just the orthogonal vector $(\cos,-\sin)$.
If you know complex numbers, this is even easier to see there, since the vector corresponds to $u_x = \cos x + i \sin x = e^{ix}$. And the definition of this exponential is exactly the function such that itsderivative is $u_x' = i u_x$, which is the vector orthogonal and of size $1$.
A: To answer this, we'd need a definition of "trigonometric", so I'd say trigonometric functions are real-valued rational functions of $e^{ix}$, whose closure rules are trivial. By contrast, logarithms of rational functions of $ix,\,\sqrt{1-x^2}$ don't achieve the same, which is why the situation is different with inverse trigonometric functions.
A: A simple way to find the derivative of $\sin{x}$ is shown below. A similar method works for $\cos{x}$ and $\tan{x}$:
$$\frac{d}{dx}\sin{x}=\lim_{h\to0}\frac{\sin{(x+h)}-\sin{x}}{h}=\lim_{h\to0}\frac{\sin{x}\cos{h}+\sin{h}\cos{x}-\sin{x}}{h}$$
$$=\lim_{h\to0}\frac{\sin{h}}{h}\cos{x}+\lim_{h\to0}(\sin{x}(\frac{\cos{h}-1}{h}))$$
I'm afraid you'll have to take my word that
$$\lim_{h\to0}\frac{\sin{h}}{h}=1,~~\lim_{h\to0}\frac{\cos{h}-1}{h}=0$$
which can easily be checked graphically eg on Desmos. So we have
$$\frac{d}{dx}\sin{x}=\lim_{h\to0}\frac{\sin{h}}{h}\cos{x}+\lim_{h\to0}(\sin{x}(\frac{\cos{h}-1}{h}))=\lim_{h\to0}\cos{x}+\lim_{h\to0}0=\cos{x}$$
as we expected.
